Integrand size = 43, antiderivative size = 26 \[ \int \frac {e^{e^x x^2} \left (-1+e^x x^2 (2+x)\right )+4 x^2 \log (\log (5))}{4 x^2 \log (\log (5))} \, dx=x+\frac {-1+\frac {e^{e^x x^2}}{x}}{4 \log (\log (5))} \]
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Leaf count is larger than twice the leaf count of optimal. \(55\) vs. \(2(26)=52\).
Time = 0.08 (sec) , antiderivative size = 55, normalized size of antiderivative = 2.12, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.070, Rules used = {12, 14, 2326} \[ \int \frac {e^{e^x x^2} \left (-1+e^x x^2 (2+x)\right )+4 x^2 \log (\log (5))}{4 x^2 \log (\log (5))} \, dx=\frac {e^{e^x x^2} \left (e^x x^3+2 e^x x^2\right )}{4 \left (e^x x^2+2 e^x x\right ) x^2 \log (\log (5))}+x \]
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Rule 12
Rule 14
Rule 2326
Rubi steps \begin{align*} \text {integral}& = \frac {\int \frac {e^{e^x x^2} \left (-1+e^x x^2 (2+x)\right )+4 x^2 \log (\log (5))}{x^2} \, dx}{4 \log (\log (5))} \\ & = \frac {\int \left (\frac {e^{e^x x^2} \left (-1+2 e^x x^2+e^x x^3\right )}{x^2}+4 \log (\log (5))\right ) \, dx}{4 \log (\log (5))} \\ & = x+\frac {\int \frac {e^{e^x x^2} \left (-1+2 e^x x^2+e^x x^3\right )}{x^2} \, dx}{4 \log (\log (5))} \\ & = x+\frac {e^{e^x x^2} \left (2 e^x x^2+e^x x^3\right )}{4 x^2 \left (2 e^x x+e^x x^2\right ) \log (\log (5))} \\ \end{align*}
Time = 0.05 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.88 \[ \int \frac {e^{e^x x^2} \left (-1+e^x x^2 (2+x)\right )+4 x^2 \log (\log (5))}{4 x^2 \log (\log (5))} \, dx=x+\frac {e^{e^x x^2}}{4 x \log (\log (5))} \]
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Time = 2.12 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.77
method | result | size |
risch | \(x +\frac {{\mathrm e}^{{\mathrm e}^{x} x^{2}}}{4 \ln \left (\ln \left (5\right )\right ) x}\) | \(20\) |
norman | \(\frac {x^{2}+\frac {{\mathrm e}^{{\mathrm e}^{\ln \left (x^{2}\right )+x}}}{4 \ln \left (\ln \left (5\right )\right )}}{x}\) | \(24\) |
default | \(\frac {4 \ln \left (\ln \left (5\right )\right ) x^{2}+{\mathrm e}^{{\mathrm e}^{\ln \left (x^{2}\right )+x}}}{4 \ln \left (\ln \left (5\right )\right ) x}\) | \(28\) |
parallelrisch | \(\frac {8 \ln \left (\ln \left (5\right )\right ) x^{2}+2 \,{\mathrm e}^{{\mathrm e}^{\ln \left (x^{2}\right )+x}}}{8 \ln \left (\ln \left (5\right )\right ) x}\) | \(30\) |
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Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.04 \[ \int \frac {e^{e^x x^2} \left (-1+e^x x^2 (2+x)\right )+4 x^2 \log (\log (5))}{4 x^2 \log (\log (5))} \, dx=\frac {4 \, x^{2} \log \left (\log \left (5\right )\right ) + e^{\left (e^{\left (x + \log \left (x^{2}\right )\right )}\right )}}{4 \, x \log \left (\log \left (5\right )\right )} \]
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Time = 0.10 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.65 \[ \int \frac {e^{e^x x^2} \left (-1+e^x x^2 (2+x)\right )+4 x^2 \log (\log (5))}{4 x^2 \log (\log (5))} \, dx=x + \frac {e^{x^{2} e^{x}}}{4 x \log {\left (\log {\left (5 \right )} \right )}} \]
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Time = 0.23 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.96 \[ \int \frac {e^{e^x x^2} \left (-1+e^x x^2 (2+x)\right )+4 x^2 \log (\log (5))}{4 x^2 \log (\log (5))} \, dx=\frac {4 \, x \log \left (\log \left (5\right )\right ) + \frac {e^{\left (x^{2} e^{x}\right )}}{x}}{4 \, \log \left (\log \left (5\right )\right )} \]
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\[ \int \frac {e^{e^x x^2} \left (-1+e^x x^2 (2+x)\right )+4 x^2 \log (\log (5))}{4 x^2 \log (\log (5))} \, dx=\int { \frac {4 \, x^{2} \log \left (\log \left (5\right )\right ) + {\left ({\left (x + 2\right )} e^{\left (x + \log \left (x^{2}\right )\right )} - 1\right )} e^{\left (e^{\left (x + \log \left (x^{2}\right )\right )}\right )}}{4 \, x^{2} \log \left (\log \left (5\right )\right )} \,d x } \]
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Time = 9.47 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.73 \[ \int \frac {e^{e^x x^2} \left (-1+e^x x^2 (2+x)\right )+4 x^2 \log (\log (5))}{4 x^2 \log (\log (5))} \, dx=x+\frac {{\mathrm {e}}^{x^2\,{\mathrm {e}}^x}}{4\,x\,\ln \left (\ln \left (5\right )\right )} \]
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